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Question

Find the equation of the parabola whose axis is parallel to $$y$$-axis which passes through points $$(4 , 5)$$,$$(-2 , 11)$$ and $$(-4 , 21)$$.


Solution

A parabola with axis parallel to y-axis has equal to $$y=ax^2+bx+c$$, for constants a, b, c in each of the three points.

$$5=a(4)^2+b(4)+c$$

$$11=a(-2)^2+b(-2)+c$$

$$21=a(-4)^2+b(-4)+c$$

So, $$\left(\begin{matrix} 16 & 4 & 1 \\ 4 & -2 & 1\\ 16 & -4 & 1\end{matrix}\right|\left.\begin{matrix} 5 \\ 11\\ 21 \end{matrix}\right)$$ is our system.

This simplifies as follows

$$\Rightarrow \left(\begin{matrix} 16 & 4 & 1\\ 4 & -2 & 1\\ 0 & -8 & 0\end{matrix}\right|\left.\begin{matrix} 5 \\ 11\\ 16\end{matrix}\right)$$

$$\Rightarrow \left(\begin{matrix} 1 & 1/4 & 1/16 \\ 0 & -3 & 3/4\\ 0 & 1 & 0\end{matrix}\right|\left.\begin{matrix} 5/16\\ 39/4\\ -2\end{matrix}\right)$$

$$\Rightarrow \left(\begin{matrix} 1 & 0 & 1/16\\ 0 & 1 & 0\\ 0 & 0 & 1\end{matrix}\right|\left.\begin{matrix} 13/16\\ -2 \\ 5\end{matrix}\right)$$

Now we get

$$c=5$$

$$b=2$$

$$a=\dfrac {13}{16}-5\dfrac{1}{16}$$

$$=\dfrac{13-5}{16}$$

$$=\dfrac{8}{16}$$

$$=\dfrac{1}{2}$$

Finally we write $$y=\dfrac{1}{2}.x^2-2x+5$$.

Maths

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